The Discrete Charm of Isometries

Table of Contents

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1. Motivation

In the lesson on reflections, we defined the charm of an
isometry to be the minimal number of mirrors required to
express the isometry as the composition of their reflections.
We showed that the charm never exceeds 4. Here we apply
what we have studied to show that charm never exceeds 3, and
identify the most charming isometries, those with charm equal to 3.

The method we shall use is that of "elimination" of all possibilities.
This requires some care that we don’t overlook a case. Many a theorem
acquired a bad proof, which overlooks an obscure case.

2. Reduction of Factors

Recall that the identity has charm 0.

If an isometry, $ \alpha = \sigma_\ell $, is a reflections then
its charm is just 1.

If an isometry, $ \alpha = \sigma_k \sigma_h $, is the product
of two reflection, the two lines $ h, k $ can be in one of three
positions

So, in all these cases the charm of $ \alpha $ is 1, and we
can assume that all three lines are different. The exactly one of three
cases occur, all three are concurrent, all three are parallel (and
so concurrent at infinity), or not.

If they are concurrent at $ Q = (fg) =(gh) $ then we can
group the three reflections to equal a reflection $ \sigma_h $
following a rotation, or a rotation following a reflection. Let us
recalibrate the latter case choosing as mirror $ f $ to replace
$ g $. Then there is an "after mirror" so that
$ \sigma_{f} \sigma_{g} = \sigma_a \sigma_f $. Substituting we get
$ \alpha =
(\sigma_a \sigma_f)\sigma_{f} = \sigma_a (\sigma_{f} \sigma_{f})
= \sigma_a $ .

Question 1.

What if we had recalibrated the other way, obtaining
$ \alpha = \sigma_b $ for an appropriate "before mirror"?

If we calculate (exercise) the two mirrors exactly, using the recalibration
theorem, the we will find that $ a = b $, as expected.

Next, suppose the three mirrors are concurrent at an ideal point, i.e. all
three are mutually parallel. Then (exercise) apply the recalibration theorem
for translations in an analogous way.

This leaves the case that the three mirrors are not concurrent in either
sense and form, a possibly ideal, triangle. These isometries are the
most charming ones requires us to settle the case that an isometry
is the product of four reflections. To reduce notational clutter we
shall use subscripts.

If two adjacent mirrors are equal, then the square of an involution
reduces the $ \alpha $ to the product of two reflections,
treated above.

If either $ m_4 = m_2 $ or $ m_3 = m_1 $ then
$ \alpha $ is the product of two reflections, one of which
is the conjugate ot $ m_3 $ or $ m_2 $ respectively.

If $ m_4 = m_1 $ then $ \alpha $ is the product of
the conjugate of $ m_3 $ and of $ m_2 $ by the same
reflection. (Exercise.)

Finally, we may assume that all four mirrors are different. Here the
use ideal centers of rotation for translations pays off. Let
$ A = (m_4 m_3) , B = (m_2 m_1)$ , whether neither, one or both
are ideal. (Exercise, draw pictures of these four cases ). Now
either $ A = B $ or not.

Suppose first that the two centers are the same. Note that means that
all four lines are either all parallel, or all concurrent. In the
parallel case we use the additive algebra of parallel vectors to
discover which translation $ \alpha $ collapses to. In the
concurrent case, we use the clock arithmetic of rotations about the
same center to discover the exact angle for the rotation $ alpha $.
(Exercises.)

This leaves the case that $ A \ne B $ and so $ m = (AB) $ is
a well defined line. (Exercise: what is $ m $ in each of the
four cases that one, both or none of the two points are ideal? )

There is one case that the previous paragraph doesn’t make senses.
Suppose both $ A \ne B $ are ideal. The $ m = \infty $
the ideal line. What does it mean to reflect in the ideal line?
But if we examine this case separately, all we’re saying is that
the first two mirrors are parallel, and the second two mirrors are
parallel, but the two sets are not parallel. Since each pair is
a translation, and we know how vectors add, we see that the product
of four reflections again collapse into a single translation.
(Exercise: Why is this translation not the identity.)

3. Preliminary Classification Theorem

What we have now proved is that an isometry is

The identity (charm 0)

A reflection (charm 1)

A rotation or translations (charm 2)

A mystery (charm 3).

Further, we can distinguish between a rotation and a translation
by counting the fixed points (1 respectively 0).

In the next and final section we discover the identity of the
the mysterious, and most charming isometries.